Expanded form strategy

If the rise and run on one section are, respectively, 3.0 m and 6.3 m, what is the run on the other section if its rise is 4.3 m

BlackZzzverrR [31]

The the run on the other section if its rise is 4.3 m = 9.03m

<h3>How top solve for the run</h3>

To get the run on the other section we have to use the cross multiplication formula, using m1 = m2

Such that we would have

3.0/6.3 = 4.3/x

6.3 * 4.3 = 3.0 * x

= 27.09 = 3x

divide through by 3

x = 9.03

Hence we conclude that the the run on the other section if its rise is 4.3 m = 9.03m

Read more on equations here: brainly.com/question/2972832

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3 0

2 years ago

26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics

hjlf

The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

$\mu _d = \mu _{post} - \mu _{pre}$

Then the null hypotheses and alternative hypotheses will be

H₀: $\mu _d = 0$

Hₐ: $\mu _d > 0$

Then the test statistic will be

$\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5$

Then

$\rm S_d = 6.02$

The test statistic value is given by

$\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16$

Since this is a right-tailed test, so the critical value is given by

$\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353$

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that $\mu _d > 0$ that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

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7 0

2 years ago

The sports shop has 57 part-time employees and 19 full-time employees. What percentage of the employees at the sports shop work

Anna [14]

The answer is 75%. To find this answer, you add 57 and 19 to get the total workers which is 76. Then, you find the percentage of 57 out of 76. 57 divided by 76 is .75, bu to get the percent, you have to move the decimal two places to the right.

4 0

3 years ago

What is T/8 = -4 solve for T

andrew11 [14]

Answer:

The answer is -32

Step-by-step explanation:

Multiply both sides by 8 to cancel 8 as the denominator and you have T = -32

7 0

4 years ago

Write a function rule for each table.

Sergeu [11.5K]

Answer: p = 7.50h

Step-by-step explanation: If you compare the hours worked (h) with pay (p) you'll find that the pay is always 7.50 times the hours worked in dollars. Hence, p = 7.50h.

5 0

4 years ago